A tube light of 60 V, 60 W rating is connected across an AC source of 100 V and 50 Hz frequency. Then,

$P_R=V_{R}I$

$I=\frac{P_R}{V_R}=\frac{60}{60}=1A$

$V_L=\sqrt{V^2-V_R^2}$

$=\sqrt{(100{)}^2-(60{)}^2}$

$=80\mathrm{V}$

$80=I\times X_L=I\left(2\pi fL\right)$

$L=\frac{80}{2\pi fI}=\frac{4}{5\mathrm{\pi }}H$

If we connect another resistance R in series, then voltage drop across it should be 40V, so that remaining 60V is used by the tube light.

$R=\frac{V}{I}=\frac{40}{1}=40\mathrm{\Omega }$